【codeforces 721A】One-dimensional Japanese Crossword

codancer

271562人浏览 · 2018-04-27 17:05:03

codancer · 2018-04-27 17:05:03 发布

Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia https://en.wikipedia.org/wiki/Japanese_crossword).

Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.

$\text{[math]}$ The example of encrypting of a single row of japanese crossword.

Help Adaltik find the numbers encrypting the row he drew.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).

Output

The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.

The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.

  Examples
 
    input 
    
     Copy
    
3
BBW

    output 
    
     Copy
    
1
2 

    input 
    
     Copy
    
5
BWBWB

    output 
    
     Copy
    
3
1 1 1 

    input 
    
     Copy
    
4
WWWW

    output 
    
     Copy
    
0

    input 
    
     Copy
    
4
BBBB

    output 
    
     Copy
    
1
4 

    input 
    
     Copy
    
13
WBBBBWWBWBBBW

    output 
    
     Copy
    
3
4 1 3

题意看的一脸懵逼，仔细看了看样例感觉应该是求整个字符串中连续的B的个数，每段输出，居然过了，有时候还要靠猜测题意了。。。

AC 代码：

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
char c[10005];
int re[10005];
int main()
{
	int n;
	cin>>n;
	memset(re,0,sizeof(re));
	for(int i=0;i<n;i++)
	{
		cin>>c[i];
	}
	int k=0,sum=0;
	for(int i=0;i<n;i++)
	{
		int a=0;
		if(c[i]=='B')
		{
			while(c[i]=='B')
			{
				a++;//连续区间内B的个数
				i++;
			}
			sum++;//连续区间的个数
			re[k++]=a;
		}
	}
	cout<<sum<<endl;
	if(sum)
	{
		for(int i=0;i<k;i++)
		{
			if(i) cout<<" "<<re[i];
			else cout<<re[i];
		}
	cout<<endl;	
	}
 }